From Ebbinghaus, Flum, and Thomas' Mathematical Logic, chapter 2, exercise 1.3:
Let be given. For
such that show that there is a point in the closed interval
such that .
Conclude from this that , and hence also, are uncountable.
(Hint: By induction define a sequence
of closed intervals such that and use the fact that
Define . For any , if ,
let and define
For any , we can see that for any value of , we have
Therefore, if , then
for any .
As there is some
, and for this , we have
(and ) and
for any , so for any given , then
is not surjective onto (and ).
Therefore, (and ) is not countable.